## College Algebra (10th Edition)

$\color{blue}{x = \left\{-3, 10\right\}}$
Since the leading coefficient is $1$, the trinomial can be factored by looking for the factors $c$ and $d$ of the constant term $-30$ whose sum is equal to the middle term's coefficient ($c+d=-7$). The factored form of the trinomial then is $(x+c)(x+d)$ Note that: $-10(3)=-30$ and $-10+3=-7$ Thus, the factors of $-30$ that we are looking for are $-10$ and $3$. This means that the factored form of the trinomial is: $=(x+(-10))(x+3) \\=(x-10(x+3)$ Therefore the equation becomes: $(x-10)(x+3)=0$ Use the Zero-Product Property (which states that if $xy=0$, then either $x=0$ or $y=0$ or both are zero) by equating each factor to zero to obtain: $x-10=0 \text{ or } x+3=0$ Solve each equation to obtain: $x=10$ or $x=-3$ Therefore, the solutions to the given equation are: $\color{blue}{x = \left\{-3, 10\right\}}$